# Simple functions in the product space

Dear all,

Let $(X,\mathcal{F},\mu)$ and $(G,\mathcal{G},\nu)$ be two measure spaces with $\mu$ and $\nu$ s$\sigma$-finite. Per definition linear combinatoins of $$\mathbb{1}_C(x,y)$$ for $C\subset X\times Y$ is dense in any $L^p(X\times Y,\mu \otimes \nu)$. This should also be true for the linear combinations of

$$\mathbb{1}_{A \times B}(x,y)$$

for $A\subset X$ and $B\subset Y$. Do you know a reference?

Thank you,

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Real Analysis: Modern techniques and their applications by G.B.Folland is a good reference. –  Uday Sep 3 '12 at 11:51

A useful density lemma is the following.

Let $(X,\mathcal{A}, \mu)$ be a measure space and let $\Gamma\subset > \mathcal{P}(X)$ a ring of sets of finite measure that generates the $\sigma$-algebra $\mathcal{A}$. Then, the linear span of the characteristic functions of sets in $\Gamma$ is dense in $L^p$ (here $1\le p < +\infty$)

In your case, of course, you can take $\Gamma$ to be the collection of finite unions of Cartesian products of measurable sets of finite measure.

Incidentally, the hypothesis on $\Gamma$ in that density lemma can be weakened (one does not need all the ring strucure of $\Gamma$). Let's say that $\Gamma\subset \mathcal{P}(X)$ is a "semi-ring" (warning: not standard; I borrowed it from Halmos, with a slightly more general meaning) if the following holds:

For all $A$ and $B$ in $\Gamma$ the sets $A\setminus B$ and $A\cap B$ are both expressible as union of countably many disjoint element of $\Gamma$.

Then the above lemma holds true. The notion of "semi-ring"is also interesting, in that it is a convenient domain for a completely additive set function, in order that the Caratheodory's Extension Theorem holds.

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Thank you, do you have a reference for the first Lemma? Best, warsaga –  warsaga Sep 3 '12 at 16:06
Usually, text on probability, e.g. Kallenberg, Foundations of Modern Probability. But you can prove that lemma on the lines of Michael Greinecker's answer, using Beppo Levi's theorem. Consider the closure $V$ of the linear span of $1_{|A}$ with $A\in\Gamma$. Then, there are in $V$ all characteristic functions $1_{|B}$ with $B$ a set of finite measure that is a countable union of elements of $\Gamma$. And then, there are in $V$ all characteristic functions $1_{|C}$ with $C$ an intersection of a countable decreasing sequence $B_k$ of the preceding form (...) –  Pietro Majer Sep 3 '12 at 17:28
.. and these sets $C$ are all the measurable subsets of finite measure, up to null measure sets. Thus $V$ has the characteristic functions of all sets of finite measure, therefore it is $L^p$ since it is closed. –  Pietro Majer Sep 3 '12 at 17:30
sorry to ask once more. Why does C contain all sets except sets of measure zero? Moreover it is only clear that C contains disjoint countable union. Thank you? –  warsaga Sep 5 '12 at 11:14
I mean: any measurable set $S$ of finite measure is of the form $C$ up to a perturbation of measure zero, that is, for any such $S$ there is a set $C$ such that $S\Delta C=0$ (that is $1_C=1_S$ a.e.) –  Pietro Majer Sep 6 '12 at 20:29
I know a result that the indicator functions of product measurable sets can be approximated by indicator functions of one variable. That is, for every $A\in F*G$(product sigma algebra), there exist $F_k\in F$ and $G_k\in G$ such that \sum_{k=1}^n 1_{F_k}1_{G_k} -->1_{A}. But I don not know how to prove it. Do you know a reference or give me some helps? By the way, can the convergence be monotonous? And can any measurable function $f$ defined on $X\times Y$ be approximated by linear combination of indicator functions of one variables monotonically?